Question

Find the electric field at point $P$ as shown in the figure on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q$. The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2}L$.

• A. $\frac{Q}{4\pi {ϵ}_0{L}^2}$
• B. $\frac{Q}{3\pi {ϵ}_0{L}^2}$
• C. $\frac{\sqrt{3}Q}{4\pi {ϵ}_0{L}^2}$
• D. $\frac{Q}{2\sqrt{3}\pi {ϵ}_0{L}^2}$

Answer 1

The correct option is D $\frac{Q}{2\sqrt{3}\pi {ϵ}_0{L}^2}$


$\tan \theta =\frac{L/2}{a}=\frac{L/2}{\sqrt{3}L/2}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={30}^{\circ }$

Now,

$E_{net}=\frac{K\lambda }{a}[\sin {\theta }_1+\sin {\theta }_2]$

Here, ${\theta }_1={\theta }_2={30}^{\circ }$ and $\lambda =\frac{Q}{L}$

So, $E_{net}=\frac{Q}{4\pi {ϵ}_{0}L\times \left(\frac{\sqrt{3}}{2}L\right)}[\sin {30}^{\circ }+\sin {30}^{\circ }]$

$=\frac{Q}{2\sqrt{3}\pi {ϵ}_0{L}^2}[\frac{1}{2}+\frac{1}{2}]$

$=\frac{Q}{2\sqrt{3}\pi {ϵ}_0{L}^2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.